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 5.1 Math IC Algebra Strategies 5.2 Writing Equations 5.3 Manipulating Equations 5.4 Zero Product 5.5 Absolute Value 5.6 Inequalities 5.7 Systems of Equations

 5.8 Common Word Problems 5.9 Logarithms 5.10 Polynomials 5.11 Key Formulas 5.12 Review Questions 5.13 Explanations
Polynomials
A polynomial is an expression that contains one or more algebraic terms, each consisting of a constant multiplied by a variable raised to a power greater than or equal to zero. For example, is a polynomial with three terms (the third term is . , on the other hand, is not a polynomial because x is raised to a negative power. A binomial is a polynomial with exactly two terms: and are both binomials.
The rest of this chapter will show you how to perform different operations on and with polynomials.
Multiplying Binomials
There is a very simple acronym that is useful in remembering how to multiply binomials. It is FOIL, and it stands for First, Outer, Inner, Last. This is the order that you multiply the terms of two binomials to get the right product.
For example, if asked to multiply the binomials:
You first multiply the first terms of each binomial:
Next, multiply the outer terms of the binomials:
Then, multiply the inner terms:
Finally, multiply the last terms:
Combine like terms and you have your product:
Here are a few more examples:
Multiplying Polynomials
Every once in a while, the Math IC test will ask you to multiply polynomials. It may seem like a daunting task. But when the process is broken down, multiplying polynomials requires nothing more than distribution and combining like terms.
Consider the polynomials (a + b + c) and (d + e + f). To find their product, just distribute the terms of the first polynomial into the second polynomial individually and combine like terms to formulate your final answer:
Here’s another example:
As you can see, multiplying polynomials is little more than rote multiplication and addition.
A quadratic, or quadratic polynomial, is a polynomial of the form ax2 + bx + c, where a ≠ 0. The following polynomials are quadratics:
A quadratic equation sets a quadratic polynomial equal to zero. That is, a quadratic equation is an equation of the form ax2 + bx + c = 0. The values of x for which the equation holds are called the roots, or solutions, of the quadratic equation. Most of the questions on quadratic equations involve finding their roots.
There are two basic ways to find roots: by factoring and by using the quadratic fo-rmula. Factoring is faster, but it can’t always be done. The quadratic formula takes longer to work out, but it works for all quadratic equations. We’ll study both in detail.
Factoring
To factor a quadratic, you must express it as the product of two binomials. In essence, factoring a quadratic involves a reverse-FOIL process. Take a look at this quadratic:
In the example above, the leading term has a coefficient of 1 (since 1x2 is the same as x2). Since the two x variables are multiplied together during the FIRST step of foiling to get the first term of the quadratic polynomial, we know that the binomials whose product is this quadratic must be of the form (x + m)(x + n), where m and n are constants. You also know that the sum of m and n is 10, since the 10x is derived from multiplying the OUTER and INNER terms of the binomials and then adding the resulting terms together (10x = mx + nx, so m + n must equal 10). Finally, you know that the product of m and n equals 21, since 21 is the product of the two last terms of the binomials.
Now you just need to put the pieces together to find the values of m and n. You know that x is the first term of both binomials, and you know that the sum of m and n is 10 and the product of m and n is 21. The pair of numbers that fit the bill for m and n are 3 and 7. Thus, x2 + 10x + 21 = (x + 3)(x + 7). The quadratic expression has now been factored and simplified.
On the Math IC, though, you will often be presented with a quadratic equation. The only difference between a quadratic equation and a quadratic expression is that the equation is set equal to 0 (x2 + 10x + 21 = 0). If you have such an equation, then once you have factored the quadratic you can solve it. Because the product of two terms is zero, one of the terms must be equal to zero. Thus, since x + 3 = 0 or x + 7 = 0, the solutions (also known as the roots) of the quadratic must be x = –3 and x = –7.
So far we’ve dealt only with quadratics in which the terms are all positive. Factoring a quadratic that has negative terms is no more difficult, but it might take slightly longer to get the hang of it, simply because you are less used to thinking about negative numbers.
Consider the quadratic equation x2 – 4x – 21 = 0. There are a number of things you can tell from this equation: the first term of each binomial is x, since the first term of the quadratic is x2; the product of m and n is –21; and the sum of a and b equals –4. The equation also tells you that either m or n must be negative but that both cannot be negative, because the multiplication of one positive and one negative number can only result in a negative number. Now you need to look for the numbers that fit these requirements for m and n. The numbers that multiply together to give you –21 are: –21 and 1, –7 and 3, 3 and –7, and 21 and –1. The pair that works in the equation is –7 and 3.
There are two special quadratic polynomials that pop up quite frequently on the Math IC, and you should memorize them. They are the perfect square and the difference of two squares. If you memorize the formulas below, you may be able to avoid the time taken by factoring.
There are two kinds of perfect square quadratics. They are:
1. a2 + 2ab + b2 = (a + b)(a + b) = (a + b)2. Example: a2 + 6ab + 9 = (a + 3)2
2. a2 – 2ab + b2 = (ab)(ab) = (ab)2. Example: a2 – 6ab + 9 = (a –3)2
Note that when you solve for the roots of a perfect square quadratic equation, the solution for the equation (a + b)2 = 0 will be –b, while the solution for (a + b)2 = 0 will be b.
Here’s an instance where knowing the perfect square or difference of two square equations can help you:
 Solve for x: 2x2 + 20x + 50 = 0.
To solve this problem by working out the math, you would do the following:
If you got to the step where you had 2(x2 + 10x +25) = 0 and realized that you were working with a perfect square of 2(x + 5)2, you could immediately have divided out the 2 from both sides of the equation and seen that the solution to the problem is –5.
Since the ability to factor quadratics relies in large part on your ability to “read” the information in the quadratic, the best way to get good is to practice, practice, practice. Just like perfecting a jump shot, repeating the same drill over and over again will make you faster and more accurate. Take a look at the following examples and try to factor them on your own before you peek at the answers.